Vectors Calaulus 1.14

Tangent Planes and Normal Lines Tangent Planes Let z = f(x,y) be a function of two variables.  We can define a new function F(x,y,z) of three  variables by subtracting z.  This has the condition         F(x,y,z)  =  0 Now consider any curve defined parametrically by         x  =  x(t),      y  =  y(t)      z  =  z(t) We can write,         F(x(t), y(t), z(t))  =  0 Differentiating both sides with respect to t, and using the chain rule gives         Fx(x, y, z) x' + Fy(x, y, z) y' + Fz(x, y, z) z'  =  0 Notice that this is the dot product of the gradient function and the vector ,         GradF .   =  0 In particular the gradient vector is orthogonal to the tangent line of any curve on the surface.   This leads to                     Definition Let F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), then the tangent plane to F ( x, y, z ) at ( x0 , y0 , z0 ) is the plane with normal vector           Grad F(x0,y0,z0) that passes through the point (x0,y0,z0).  In Particular the equation of the tangent plane is        Grad F(x0,y0,z0) . < x - x0 , y - y0 , z - z0 >  =  0 Example Find the equation of the tangent plane to         z  =  3x2 - xy at the point (1,2,1) Solution We let         F(x,y,z)  =  3x2 - xy - z then         Grad F  =  <6x - y, -x, -1> At the point (1,2,1), the normal vector is         Grad F(1,2,1) = <4, -1, -1> Now use the point normal formula for a plan         <4, -1, -1> .   =  0 or         4(x - 1) - (y - 2) - (z - 1)  =  0 Finally we get         4x - y - z  =  1 Normal Lines Given a vector and a point, there is a unique line parallel to that vector that passes through the point.  In the context of surfaces, we have the gradient vector of the surface at a given point. This leads to the following definition.                         Definition Let F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), then the normal line to F(x,y,z) at ( x0 , y0 , z0 ) is the line with normal vector           GradF(x0,y0,z0) that passes through the point (x0,y0,z0).  In Particular the equation of the normal line is           x(t) = x0 + Fx(x0,y0,z0) t            y(t) = y0 + Fy(x0,y0,z0) t            z(t) = z0 + Fz(x0,y0,z0) t Example Find the parametric equations for the normal line to         x2yz - y + z - 7  = 0 at the point (1,2,3). Solution We compute the  gradient         Grad F  =  <2xyz, x2z - 1, x2y + 1>  =  <12, 2, 3> Now use the formula to find         x(t) = 1 + 12t      y(t) = 2 + 2t      z(t) = 3 + 3t The diagram below displays the surface and the normal line.         Angle of Inclination Given a plane with normal vector n the angle of inclination, q is defined by                  |n . k|     cosq =                                  ||n||  More generally, if         F(x,y,z)  =  0 is a surface, than the angle of inclination at the point  (x0, y0, z0) is defined by the angle of inclination of the tangent plane at the point.                     |Grad F(x0, y0, z0) . k|     cosq =                                                                  ||Grad F(x0, y0, z0)|| Example Find the angle of inclination of          x2             y2          z2                  +            +             =  1           4            4            8 at the point (1,1,2). Solution First compute         Grad F  =  Now plug in to get         Grad F(1,1,2)  =  <1/2, 1/2, 1/2> We have         |<1/2, 1/2, 1/2> . k|  =  1/2 Also,         ||<1/2, 1/2, 1/2>||  =  / 2 Hence         cosq = (1/2)/[()/2] = 1/ So the angle of inclination is         q  =  arccos(1/)  @   0.955 radians The Tangent Line to a Curve Example Find the tangent line to the curve of intersection of the sphere         x2 + y 2 + z2  =  30 and the paraboloid         z  =  x2 + y2 at the point (1,2,5). Solution We find the Grad of the two surfaces at the point         Grad (x2 + y 2 + z2) = <2x, 2y, 2z> = <2, 4,10> and         Grad (x2 + y 2 - z)  =  <2x, 2y, -1> = <2, 4, -1> These two vectors will both be perpendicular to the tangent line to the curve at the point, hence their cross product will be parallel to this tangent line.  We compute         Hence the equation of the tangent line is         x(t) = 1 - 44t       y(t) = 2 + 22t       z(t) = 5